5.7 KiB
Math4121 Lecture 13
Hidden Chapter 1
This chapter is not covered in the lecture but I still want to mention it here.
At first, when the integral was first invented, it was thought to be the area under the curve or above the curve, using intuitive geometric definition from the mysterious common sense of the homo-sapiens. There was not a rigorous definition of the integral from the eighteenth century, when it was first invented, to the nineteenth century, when Riemann, Lebesgue, and others rigorously defined the integral.
The integral was thought to be the anti-derivative, for the general publics.
However, we want to apply the integral to more general functions, rather than just the differentiable functions.
So, we need a rigorous definition of the integral, one potential solution is the Cauchy-Riemann integral.
Riemann integral
Recall from the previous lectures, we have the following definition of the Riemann integral:
A function f is Riemann integrable on [a,b] if there exists a number V such that for every \epsilon>0, there exists a \delta>0 such that for every partition P=\{x_0=a,x_1,\cdots,x_n=b\} of [a,b] with mesh less than \delta, we have
\left|\sum_{i=1}^{n}f(x_i^*)(x_i-x_{i-1})-V\right|<\epsilon
where x_i^* is a point in the $i$-th subinterval [x_{i-1},x_i].
This sum only exists if the Darboux's sum defined by the following is small:
Darboux's sum
Let M_i=\sup_{x\in [x_{i-1},x_i]}f(x) and m_i=\inf_{x\in [x_{i-1},x_i]}f(x).
Then, the Darboux's sum is defined as
\underline{S}(f,P)=\sum_{i=1}^{n}m_i(x_i-x_{i-1})
and
\overline{S}(f,P)=\sum_{i=1}^{n}M_i(x_i-x_{i-1})
In this case, small means that \forall \epsilon>0, there exists a \delta>0 such that if x_i-x_{i-1}<\delta, then
\sum_{i=1}^{n}(M_i-m_i)(x_i-x_{i-1})<\epsilon
(M_i-m_i) is the oscillation of f on the $i$-th subinterval [x_{i-1},x_i].
Theorem 2.1: Riemann's Integrability Criterion (corollary version)
A function f is Riemann integrable on [a, b] if and only if for every \sigma>0 be the bound for the oscillation of f, and for any \epsilon>0, we can find a subinterval length \delta, such that for any partition P of [a, b] with each subinterval has length less than \delta, the length of the sum of the lengths of the subintervals where the oscillation exceeds \sigma is less than \epsilon.
That is, mathematically, \forall \sigma,\epsilon>0, a function f is Riemann integrable on [a, b] if there exists \delta>0 such that \forall P=\{x_0=a,x_1,\ldots,x_n=b\} where x_i-x_{i-1}<\delta.
\sum_{x_i\in J}\Delta x_i<\epsilon,\quad \text{where} \quad J=\{x_i|(M_i-m_i)>\sigma\}
Theorem 2.2: Darboux Integrability Condition
Let f be a bounded function on [a,b]. This function is Riemann integrable on [a, b] if and only if for every \epsilon > 0, there exists a partition P of [a, b] such that the upper sum \overline{S}(P;f)-\underline{S}(P;f)<\epsilon
New book Chapter 2
Riemann's motivation: Fourier series
F(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_k \sin(kx)
a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) dx
b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) dx
To study the convergence of the Fourier series, we need to study the convergence of the sequence of partial sums. (Riemann integration)
Why Riemann integration?
Let
((x)) = \begin{cases}
x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \lfloor x \rfloor + \frac{1}{2}) \\
0 & x=\lfloor x \rfloor + \frac{1}{2}\\
x-\lfloor x \rfloor - 1 & x \in (\lfloor x \rfloor + \frac{1}{2}, \lfloor x \rfloor + 1] \end{cases}
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We define
f(x) = \sum_{n=1}^{\infty} \frac{((nx))}{n^2}=\lim_{N\to\infty}\sum_{n=1}^{N} \frac{((nx))}{n^2}
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(i) The series converges uniformly over x\in[0,1].
\left|f(x)-\sum_{n=1}^{N} \frac{((nx))}{n^2}\right|\leq \sum_{n=N+1}^{\infty}\frac{|((nx))|}{n^2}\leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}<\epsilon
As a consequence, f(x)\in \mathscr{R}.
(ii) f has a discontinuity at every rational number with even denominator.
\begin{aligned}
\lim_{h\to 0^+}f(\frac{a}{2b}+h)-f(\frac{a}{2b})&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))}{n^2}-\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}))}{n^2}\\
&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&=\sum_{n=1}^{\infty}\lim_{h\to 0^+}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&>0
\end{aligned}
Back to the fundamental theorem of calculus
Suppose f is integrable on [a,b], then
F(x)=\int_a^x f(t)dt
F is continuous on [a,b].
if f is continuous at x_0, then F is differentiable at x_0 and F'(x_0)=f(x_0).
Theorem (Darboux's theorem)
If \lim_{x\to a^-}f(x)=L^-, then \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=L^-.
Proof:
h\sup_{x\in [0,h]}f(x)\geq F(a+h)\geq \inf_{x\in [0,h]}f(x)h
Consequently,
f(x)=\sum_{n=1}^{\infty} \frac{((nx))}{n^2}
then
F(x)=\int_0^x f(t)dt
is continuous on [0,1].
However, since \lim_{x\to 0^+}f(x)\neq \lim_{x\to 0^-}f(x) holds for all the rational numbers with even denominator, F is not differentiable at all the rational numbers with even denominator.
Moral: There exists a continuous function on [0,1] that is not differentiable at any rational number with even denominator. (Dense set)
Weierstrass function
g(x)=\sum_{n=0}^{\infty} a^n \cos(b^n \pi x)
where 0<a<1 and ab>1+\frac{3}{2}\pi.
g(x) is continuous on \mathbb{R} but nowhere differentiable.
If we change our integral, will be differentiable at most points?